In a Young's double slit experiment,the central point on the screen is

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900 \text{ Å}$ was $1.2 \text{ mm}$ for parallel narrow slits placed $2 \text{ mm}$ apart. In this arrangement, if the slit separation is increased by one-and-half times the previous value, then the fringe width is (in $ \text{ mm}$)

In a Young's experiment,two coherent sources are placed $0.90 \, mm$ apart and the fringes are observed one metre away. If it produces the second dark fringe at a distance of $1 \, mm$ from the central fringe,the wavelength of monochromatic light used would be

In the double slit experiment,when a glass plate (refractive index $\mu = 1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$),the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is

In a Young's double slit experiment,slits are separated by $0.5 \ mm$,and the screen is placed $150 \ cm$ away. $A$ beam of light consisting of two wavelengths,$650 \ nm$ and $520 \ nm$,is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is...... $mm$.

In Young's experiment,the distance between two slits is $d/3$ and the distance between the screen and the slits is $3D$. The number of fringes in $1/3 \ m$ on the screen,formed by monochromatic light of wavelength $3\lambda$,will be